2305By AM GM inequality we have,
(ap)2 + b2 ≥ 2abp
(bp)2 + c2 ≥ 2cbp
(cp)2 + d2 ≥ 2cpd
Therefore,
(ap)2 + b2+ (bp)2 + c2 +(cp)2 + d2 - 2cpd - 2cbp - 2abp ≥ 0
Since the above condition is given as RHS≤0
(ap)2 + b2+ (bp)2 + c2 +(cp)2 + d2 - 2cpd - 2cbp - 2abp = 0
Equality holds when all the 3 inequalities are equal.
This is only possible when ap = b, bp = c and cp = d
Or,ba = cb = cd=p
Hence they are in GP.
